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Set $A=\{\,x \in\mathbb R : x^2-(a+b)x +ab=0\,\}$. We have to express it using tabular method.

At first, we factorize it, getting $(x-a)(x-b)=0$ or $x =a$ and $x=b$. But how do we know that $a$ and $b$ belongs to $\mathbb R$? Is there any proof of this fact, or are we just assuming it?

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Can you tell us a little about "tabular form"? –  Shaun Ault Jul 24 '13 at 12:51
    
@Shain Ault,tabular method,I meant.It is a form of expressing a set simply by listing out all the elements. –  rah4927 Jul 24 '13 at 13:01

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Unless it is given (e.g. in the problem statement) that $a,b\in\mathbb R$, you can't assume it. So if $a=1+i\notin \mathbb R$ and $b=7\in \mathbb R$, your tabular method should yield $\{7\}$, not $\{1+i,7\}$. If nothing about $a,b$ is specified, you'd have to answer with listing various cases (according to if/which/how many of $a,b$ are real).

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Since $A=\{\,x \in\mathbb R : x^2-(a+b)x +ab=0\,\}$, $x \in\mathbb R$ which means that both $a+b$ and $a*b$ must also be real. Implying that either $a$ and $b$ are either both real, or both complex (namely $b$ is $a$'s conjugate)

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